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CHRISTINE BREINER: Welcome
back to recitation.

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In this video I would like
us to use Green's theorem

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to compute the
following integral,

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where it's the integral
over the curve C,

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where C is the
circle drawn here.

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So the circle is oriented so
the interior is on the left

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and it's centered at the point
where x equals a, y equals 0.

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So the integral-- you can read
it but I will read it also

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for you-- it's the integral of 3
x squared y squared dx plus 2 x

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squared times the
quantity 1 plus x*y dy.

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So you are supposed
to use Green's theorem

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to compute that.

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And why don't you
pause the video,

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give it a shot, and then when
you're ready to see my solution

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you can bring the video back up.

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OK, welcome back.

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Well, what we're
going to do, obviously

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to use Green's theorem
to compute this--

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because I gave you a big
hint that you're supposed

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to use Green's
theorem, because we

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have an integral
over a closed curve.

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And what we're
going to do is now

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take the integral of another
certain integrand-- obviously

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related in some way to
this-- over the region

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that I will shade here.

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So we're interested in
this shaded region now.

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So let me write down what
Green's theorem is and then

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we'll put in the
important parts.

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OK, so just to remind you, I'm
not going to write down all

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the hypotheses of Green's
theorem that we need,

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but the point that I want
to make is that if we start

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the integral over this closed
curve C of M*dx plus N*dy we

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can also integrate over the
region that C encloses of N sub

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x minus M sub y dy dx.

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So that is ultimately
what we're going to do.

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And so in this case
again, as always,

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M is going to be the function
associated with the dx portion.

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It's the i-th component
of the vector field.

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And N is going to be the
function associated here

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with the dy.

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That's standard obviously.

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So now what we want
to do is transform

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what we have there into
something that looks like this.

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So the region-- I'm just
going to keep calling it

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R for the moment-- but the
region R, you'll notice,

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is the thing that I
shaded in the drawing.

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And so now let's compute
what this is-- well

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you have the
capacity to do that.

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That's just some straight
taking derivative.

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So I'm just going to write down
what it is and not show you

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all the individual pieces.

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So I'll just write
down what you get

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and then the simplified version.

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So you get 6 x squared
y plus 4x is N sub x.

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And then M sub y is
negative 6 x squared y.

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And I'm going to call dy*dx, I'm
just going to refer to it now

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as dA, because that makes
it a lot easier to write.

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Oh, I guess I should call this--
well, dA, this is the area.

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The volume form there.

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So this simplifies and I
just get the integral of 4x

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over this region dA.

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Now this-- you saw
an example of this

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in lecture of how to deal
with these types of problems.

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At this point, if
I really wanted to,

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I could figure out what the
bounds are in that region

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R in terms of x and y, and
I could do a lot of work

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and integrate it all, or I
could remember one simple fact.

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Which is that if
I have-- I think

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you see it in class as x
bar-- the center of mass

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should be equal to
1 over the volume

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of the region times the integral
of x dA over the region.

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That's what we know.

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This should be-- this maybe
doesn't look like a V.

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But so in this case,
volume is area, isn't it?

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Maybe I should write area,
that might make you nervous.

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So if I take the area--
I could just say A of R--

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if I take the area,
1 over the area,

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and then I multiply by the
integral of x over R with

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dy*dx, with respect to dA,
then I get the center of mass.

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Well, let's look at what--
in this picture, what

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is the center of mass here?

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If I want to balance this
thing on a pencil tip,

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if I want to balance this
disc-- assuming the density is

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everywhere the same--
on a pencil tip,

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where am I going
to put the pencil?

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I'm going to put it
right at the center.

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The x-value there is a,
the y-value there is 0.

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So if I had the y center
of mass, I would want 0.

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But I want the x center
of mass, so I want a.

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This x bar is actually
equal to, from the picture,

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is equal to a.

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And now let's notice
what I've done.

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I have taken-- I had
this quantity here-- I've

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taken this quantity
except for the 4

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and I have a way of
writing explicitly

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what this quantity
is without actually

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doing any of the integration.

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I haven't done any sophisticated
things at this point

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except know what the
center of mass is.

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So also what is the
area of the region?

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I'm going to need that.

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The area of the region--
it's a circle of radius a.

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So the area is pi a squared.

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So this quantity
is pi a squared.

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And all this together tells me
that the integral of x over R

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dA, if I solve for this part,
I get a times pi a squared.

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So I get pi a cubed.

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And that only differs from
our answer by one thing.

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There's a scalar--
you multiply by 4

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and that gives us what we want.

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I said differs from our answer.

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Differs from what we
want by one thing.

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We just multiplied by 4 there,
so we multiply by 4 here.

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So in fact-- maybe there looks
like there was a little magic

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here, so let me point out some
of the key points at the end.

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I started off knowing I
was trying to integrate 4x

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over the region R.
And R in this case

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was a circle centered at
(a, 0) and of radius a.

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And then I said, well, I
don't want to do a lot of work

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for this.

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So I'm going to
not cheat, but I'm

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going to use my knowledge
of the center of mass

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to make this easier.

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So the center of
mass is equal to 1

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divided by the area of the
region times the integral of x

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over the region.

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So I want to find the
integral of x over the region,

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I just solve for the integral
of x over the region.

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I just solve for that part.

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The center of mass-- from
just looking at the picture

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and understanding what
the center of mass

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means-- the center of mass
in the x component is a.

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The area is pi a squared.

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So I end up with
a pi a cubed when

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I solve for the integral of x.

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And then because I wanted
the integral of 4x,

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I just multiply by 4.

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And so the final answer
is actually 4 pi a cubed.

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So what I was trying
to find, if you

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remember, was I was
trying to find the value,

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when I integrated
over this curve,

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of a certain vector field.

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And that one was going to be a
little messy to do it that way.

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But Green's theorem,
actually there's

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a lot of cancellation
which makes it much easier,

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and then the center of
mass is a nice little trick

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to use at the end.

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And then the calculation
is quite simple.

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So I think that's
where I'll stop.

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